\(The\ planes: 2x-y+4z=5 \ and \ 5x-2.5y+10z=6\ are\)
passes through \((0,0,\frac 54)\)
The equations of the planes are
\(2x - y + 4z = 5\) \(.....(1)\)
\(5x - 2.5y + 10z = 6\) \(.....(2)\)
It can be seen that,
\(\frac {a_1}{a_2} =\frac {2}{5}\)
\(\frac {b_1}{b_2}=\frac {-1}{-2.5}=\frac {2}{5 }\)
\(\frac {c_1}{c_2} =\frac {4}{10} = \frac {2}{5}\)
\(∴\frac {a_1}{a_2}=\frac {b_1}{b_2}=\frac {c_1}{c_2}\)
Therefore, the given planes are parallel.
Hence, the correct answer is B.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)\)
and \(\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)\)
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k\) and
\(\overrightarrow r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k\)
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0,0,0) 3x-4y+12z=3
(b) (3,-2,1) 2x-y+2z+3=0
(c) (2,3,-5) x+2y-2z=9
(d) (-6,0,0) 2x-3y+6z-2=0
determine whether the given planes are parallel or perpendicular,and in case they are neither, find the angles between them. (a)7x+5y+6z+30=0 and 3x-y-10z+4=0
(b)2x+y+3z-2=0 and x-2y+5=0
(c)2x-2y+4z+5=0 and 3x-3y+6z-1=0
(d)2x-y+3z-1=0 and 2x-y+3z+3=0
(e)4x+8y+z-8=0 and y+z-4=0